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Arisa [49]
3 years ago
13

A proton with an initial speed of 8.20×105 m/s is brought to rest by an electric field. Part APart complete Did the proton move

into a region of higher potential or lower potential? higher potential lower potential Previous Answers Correct Part BPart complete What was the potential difference that stopped the proton? Express your answer with the appropriate units.
Physics
1 answer:
Anastaziya [24]3 years ago
5 0

Answer:

(a) Region of higher potential

(b) ΔV=1.911V

Explanation:

For Part (a)

Proton moves into a region of higher potential because it must goes in opposite direction of force to get rest

For Part (b)

Potential difference we can calculate bu definition of kinetic energy

ΔK=mv²/2= -qΔV

express in ΔV

ΔV= -mv²/2q

Substitute the given values

So

ΔV= -mv²/2q

=-\frac{mv^2}{2e}\\ =\frac{9.11*10^{-31}*(8.20*10^{5}m/s)^2}{2*1.602*10^{-19}}\\ =1.911V

ΔV=1.911V

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You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If
kakasveta [241]

Answer:

A) for leftmost point the coordinate is -0.28m that means it should be 0.28m towards the right.

B) for rightmost case the coordinate is 0.28m which is where komila should sit.

Explanation:

Detailed calculation and explanation is shown in the image below

5 0
3 years ago
An electron is moving at 7.4x105 m/s perpendicular to a magnetic field. It experiences a force of 2.0x10–13 N. What is the magne
alisha [4.7K]

Answer:

1.69 T

Explanation:

Applying,

F = BvqsinФ.................. Equation 1

Where F = Force, B = magnetic field, v = velocity, q = charge on an electron, Ф = angle between the electron and the field.

make B the subject of the equation,

B = F/(vqsinФ)............. Equation 2

From the question,

Given: F = 2.0×10⁻¹³ N, v = 7.4×10⁵ m/s, Ф = 90°

Constant: q = 1.60×10⁻¹⁹ C

Substitute into equation 2

B =  2.0×10⁻¹³/(7.4×10⁵×1.60×10⁻¹⁹×sin90°)

B = 0.169×10

B = 1.69 T

4 0
3 years ago
How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00
klio [65]

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is 9 \times 10^{-4} m^{2}. As the capacitance of capacitor is given as follows.

            C = \frac{\epsilon_{o}A}{d}

It is known that the dielectric strength of air is as follows.

               E = 3 \times 10^{6} V/m

Expression for maximum potential difference is that the capacitor can with stand is as follows.

                       dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

               Q = CV

                   = \frac{\epsilon_{o} A}{d} \times E \times d

                   = \epsilon_{o}AE

                   = 8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}

                   = 1.062 \times 10^{-8} C

or,                = 10.62 nC

Thus, we can conclude that charge on capacitor is 10.62 nC.

5 0
3 years ago
A skydiver jumped out of a plane, determine distance he distance he descended after 5 seconds. Gravity is pulling him down at 10
Bond [772]

The distance travelled in 5 seconds is 125 m

Explanation:

The motion of the skydiver is a free fall motion, since he is acted upon the force of gravity only. Therefore, it is a uniformly accelerated motion, with constant acceleration downward, and we can find the distance he travels by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time

a is the acceleration

In this problem, we have:

u = 0 (the skydiver jumps from rest)

a=g=10 m/s^2 (acceleration of gravity)

And substituting

t = 5.0 s

we find the distance travelled:

s=0+\frac{1}{2}(10)(5)^2=125 m

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

8 0
3 years ago
A fish swims to a depth of 50.00 meters in the ocean. Assuming the density of sea water is 1.0251.025 g·cm^{-3}g⋅cm −3 , calcula
LenKa [72]

Answer:

The fish is experiencing a water pressure of 502.8 kPa.

Explanation:

The water pressure the fish is experiencing can be found as follows:

P = \rho gh  (1)

Where:

g: is the gravity = 9.81 m/s²

h: is the height (depth) = 50.0 m

ρ: is the seawater's density = 1.025 g/cm³  

By replacing the above values into equation (1) we have:

P = \rho gh = 1.025 \frac{g}{cm^{3}}*\frac{1 kg}{1000 g}*\frac{(100cm)^{3}}{1 m^{3}}*9.81 m/s^{2}*50.0 m = 502.8 kPa        

Therefore, the fish is experiencing a water pressure of 502.8 kPa.

I hope it helps you!        

8 0
2 years ago
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